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  • 19.2.7 [LeetCode 50] Pow(x, n)

    Implement pow(xn), which calculates x raised to the power n (xn).

    Example 1:

    Input: 2.00000, 10
    Output: 1024.00000
    

    Example 2:

    Input: 2.10000, 3
    Output: 9.26100
    

    Example 3:

    Input: 2.00000, -2
    Output: 0.25000
    Explanation: 2-2 = 1/22 = 1/4 = 0.25
    

    Note:

    • -100.0 < x < 100.0
    • n is a 32-bit signed integer, within the range [−231, 231 − 1]
     1 class Solution {
     2 public:
     3     double myPow(double x, long n) {
     4         if (n == 0)return 1;
     5         int _n = n;
     6         if (n < 0)n = -n;
     7         double pow = myPow(x, n / 2), ans;
     8         if (n % 2)
     9             ans = pow * pow*x;
    10         else
    11             ans=pow * pow;
    12         if (_n < 0)
    13             return 1 / ans;
    14         return ans;
    15     }
    16 };
    View Code

    有点妙哦,但还是很慢

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  • 原文地址:https://www.cnblogs.com/yalphait/p/10355471.html
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